# Does uniform continuity of |f| imply uniform continuity of f?

• B
• Eclair_de_XII
In summary: When I look at this, I need to break it down into cases which is what I did initially before I saw the following replies. First off, I want to look at ##f(x)=1## if ##x<a## and ##f(x)=-1## if ##x\geq a##:##|f|(x)=1## for all ##x##, so it is uniformly continuous, but ##f## is not even continuous much less uniformly.So, I broke it down into cases, as I said before. I've been thinking about it a little more and I don't see how to explain it any other way than just referencing the counterexample. I don't see how to just explain it in words without a
Eclair_de_XII
TL;DR Summary
Define a function ##f:D\longrightarrow \mathbb{R}## such that for some ##a\in D##, ##f(x)>0## if ##x>a## and ##f(x)\leq 0## if ##x\leq a##. Note: ##f(a)=0## by the IVT. If ##|f|## is uniformly continuous on ##D##, then does that mean ##f## is also uniformly continuous?
I'd say yes, it is. Suppose ##|f|## is uniformly continuous on ##D##.

Then for all ##\epsilon>0## there is ##\delta>0## (call this ##\delta'##) such that if ##x,y\in D##, then ##||f(x)|-|f(y)||<\epsilon##.

Define sets:

##D^+=\{x\in D: x>a\}##

##D^-=\{x\in D: x<a\}##

Restrict the domain of ##f## to ##D^+##; note that ##f=|f|## here so it follows that ##f## is uniformly continuous here. In other words, if you want the distance between the images any two points ##x,y\in D^+## to be as small as possible, you need only ensure that the distance between ##x,y## is at most ##\delta'##.

Now restrict the domain of ##f## to ##D^-##. Let ##\epsilon>0## and choose ##\delta'##. Note that if ##x,y\in D^-##, then:

\begin{align}
|f(x)-f(y)|&=&|-|f(x)|-(-|f(y))|\\
&=&|-|f(x)|+|f(y)|| \\
&<&\epsilon
\end{align}

since ##|f|## is uniformly continuous.

Finally, note that ##|f|## is continuous at ##x=a## and moreover, to ensure that the image of a point ##x## is within ##\epsilon## of ##f(a)=0##, you can choose an ##x## within ##(a-\delta',a+\delta')##, since ##|f|## is uniformly continuous.

Let ##\epsilon>0## and choose ##\delta=\delta'>0##. Suppose ##x\in D^+ \cup D^-##. Then:

\begin{align}
|f(x)-f(a)|&=&|f(x)|\\
&=&||f|(x)|\\
&<&\epsilon
\end{align}

Does it even imply continuity?
\begin{align*}
1\text{____________________________}&\\
&\text{_______________________________________}-1
\end{align*}

Eclair_de_XII
Oh! So ##f(x)=1## if ##x<a## and ##f(x)=-1## if ##x\geq a##.

##|f|(x)=1## for all ##x##, so it is uniformly continuous, but ##f## is not even continuous much less uniformly.

No, it doesn't. Thanks.

I think I made a mistake when I assumed that ##f## was continuous when citing the IVT.

Eclair_de_XII said:
Note: ##f(a)=0## by the IVT.

Last edited:
fresh_42 said:
Does it even imply continuity?
\begin{align*}
1\text{____________________________}&\\
&\text{_______________________________________}-1
\end{align*}
But does this satisfy the stipulation that |f| is uniformly continuous at the point x=a, where f(a)=0? Does D omit a?

FactChecker said:
But does this satisfy the stipulation that |f| is uniformly continuous at the point x=a, where f(a)=0? Does D omit a?
Seems I only answered to the title. If ##f(a)=0,## and ##|f|## is uniformly continuous at ##a##, then it there probably won't be a significant difference between ##f## and ##|f|## at ##a##, since all terms are only locally defined.

FactChecker said:
But does this satisfy the stipulation that |f| is uniformly continuous at the point x=a, where f(a)=0? Does D omit a?

The problem statement is confused. Consider the first sentence:

Eclair_de_XII said:
Summary:: Define a function ##f:D\longrightarrow \mathbb{R}## such that for some ##a\in D##, ##f(x)>0## if ##x>a## and ##f(x)\leq 0## if ##x\leq a##.

Here there is no requirement that $f$ be continuous, or that $f(a) = 0$. A function like $$g : x \mapsto \begin{cases} -1 & x \leq a \\ 1 & x > a \end{cases}$$ satisfies the given condition. Note that $|g| = 1$ is uniformly continuous, but $g$ itself is not continuous at $a$.

But then it is said that

Note: ##f(a)=0## by the IVT.

This implies that $f$ is continuous, since if it weren't the IVT simply would not apply. I think the OP needs to clarify whether this is actually part of the problem statement or whether it is a conclusion they have arriced at themselves, and if so their justification for it.

If ##|f|## is uniformly continuous on ##D##, then does that mean ##f## is also uniformly continuous?

As the example above shows, you also require continuity of $f$, which was not expressly stated to be the case.

The condition does imply that for $x \leq a$ we have $f = -|f| \leq 0$ which is uniformly continuous if $|f|$ is, and for $x > a$ we have $f = |f| > 0$ which again is uniformly continuous if $|f|$ is. (If $|f|$ is uniformly continuous on $D$ then it is uniformly continuous on any subset of $D$: the same $\delta$ will work as for all of $D$.)

If $f$ is actually restricted to be continuous then by the above argument it is indeed uniformly continuous on $D$.

pasmith said:
OP needs to clarify whether this is actually part of the problem statement

Sorry. That was sort of an assumption that I made on my own. It was not actually part of the assignment.

If ##f## is uniformly continuous, does it follow that ##|f|## is also uniformly continuous? If ##|f|## is uniformly continuous, does it follow that ##f## is uniformly continuous? Answer the same questions with uniformly continuous replaced with ''continuous''. Explain why.

After the preceding discussion, what are your answers to the exact assignment questions?

FactChecker said:
After the preceding discussion, what are your answers to the exact assignment questions?
If ##f(a)=0## then ##||f(x)|-|f(a)||=|f(x)-f(a)|##.

fresh_42 said:
If ##f(a)=0## then ##||f(x)|-|f(a)||=|f(x)-f(a)|##.
I was meaning to ask the OP how he would answer the actual problem as stated in post #7. It is quite different.

FactChecker said:
answers to the exact assignment questions

Yes, uniform continuity of ##f## implies uniform continuity of ##|f|##. If ##\epsilon>0##, then for some ##\delta>0##, if ##x,y\in D##, then ##|x-y|<\delta## implies ##||f(x)|-|f(y)||\leq |f(x)-f(y)|<\epsilon##.

Conversely, no. Let:

##
f(x)=
##
% (Anyone know how to typeset so that this looks cleaner?)

This is the counterexample provided by Mr. fresh_42.Yes, ##|f|## is continuous for the same reasons as stated above.
Same counterexample as above.

This isn't exactly how I'm going to write it out, though.

FactChecker

## 1. What is uniform continuity?

Uniform continuity is a mathematical concept that describes the behavior of a function. A function is considered uniformly continuous if, for any given value of x, the difference between the function's output at that value and its output at a nearby value can be made arbitrarily small by choosing a small enough interval around x.

## 2. How is uniform continuity different from continuity?

Continuity and uniform continuity are both concepts that describe how a function behaves. However, continuity only requires that small changes in the input result in small changes in the output. Uniform continuity, on the other hand, requires that small changes in the input result in arbitrarily small changes in the output, regardless of where the input is on the function's domain.

## 3. Does uniform continuity of a function imply uniform continuity of its absolute value?

Yes, if a function is uniformly continuous, then its absolute value is also uniformly continuous. This is because the absolute value function is a continuous function, and the composition of two continuous functions is also continuous.

## 4. What is the significance of uniform continuity of |f| implying uniform continuity of f?

This implication is significant because it allows us to prove the uniform continuity of a function by only looking at the behavior of its absolute value. This can make proving uniform continuity easier in some cases, as the absolute value function is simpler than many other functions.

## 5. Are there any functions that are uniformly continuous but not continuous?

No, if a function is uniformly continuous, it must also be continuous. This is because uniform continuity is a stronger condition than continuity, so any function that satisfies the criteria for uniform continuity must also satisfy the criteria for continuity.

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